Quadratic Equation

17 Pages·2016·1.69 MB·English

fo/u fopkjr Hkh# tu] ugha vkjEHks dke] foifr ns[k NksM+s rqjar e/;e eu dj ';keA
iq#"k flag ladYi dj] lgrs foifr vusd] ^cuk^ u NksM+s /;s; dks] j?kqcj jk[ks VsdAA
jfpr% ekuo /keZ iz.ksrk
ln~xq# Jh j.kNksM+nklth egkjkt
STUDY PACKAGE
Subject : Mathematics
Topic : QUADRATIC EQUATIONS
Available Online : www.MathsBySuhag.com
R
Index
1. Theory
2. Short Revision
3. Exercise (Ex. 1 + 5 = 6)
4. Assertion & Reason
5. Que. from Compt. Exams
6. 39 Yrs. Que. from IIT-JEE(Advanced)
7. 15 Yrs. Que. from AIEEE (JEE Main)
Student’s Name :______________________
Class :______________________
Roll No. :______________________
Address : Plot No. 27, III- Floor, Near Patidar Studio,
Above Bond Classes, Zone-2, M.P. NAGAR, Bhopal

: 0 903 903 7779, 98930 58881, WhatsApp [phone]
www.TekoClasses.com www.MathsBySuhag.com
Get Solution of These Packages & Learn by Video Tutorials on www.MathsBySuhag.com
Quadratic Equation
1. Equation v/s Identity:
m
A quadratic equation is satisfied by exactly two values of 'x' which may be real or imaginary. The equation,
o
ax2 + bx + c = 0 is:
c 3
. a quadratic equation if a 0 Two Roots 2
g a linear equation if a = 0, b 0 One Root f
a o
h a contradiction if a = b = 0, c  0 No Root 2
u an identity if a = b = c = 0 Infinite Roots e
S If a quadratic equation is satisfied by three distinct values of 'x', then it is an identity. g
ySolved Example # 1: (i) 3x2 + 2x – 1 = 0 is a quadratic equation here a = 3. a
p
sBSolution(ii.):Here (hxi g+h 1e)s2t =p oxw2 e+r 2oxf x+ i n1 tihse a gni videenn rteitlya tiino nx .is 2 and this relation is satisfied by three different values x= 0, x
h = 1 and x = – 1 and hence it is an identity because a polynomial equation of nth degree cannot have more than n1 .
at distinct roots. 88
M2. Relation Between Roots & Co-efficients: 8
5
w. (i) The solutions of quadratic equation, ax2 + bx + c = 0, (a 0) is given by 0
3
w b  b2 4ac 9
w x = 8
2a 9
m & (ii) TIfhe, ex aprree tshseio rno,o bts2 o f4 q auacdr aDt iics ecqaullaetdio dnis,criminant of qauxa2d +ra bticx e+q cu a=t i0o,n a.  0. Then: 9, 0
7
o b c D 7
c (a)  +  =  (b) = (c) = 7
s. a a a 3
e (iii) A quadratic equation whose roots are &, is (x ) (x ) = 0 i.e. 0
9
s x2 (sum of roots) x + (product of roots) = 0 3
sSolved Example # 2: If  and are the roots of ax2 + bx + c = 0, find the equation whose roots are+2 and+2. 0
a 9
ClSolutiona(.x – 2)R2 +e pbl(axc i–n g2 )x + b yc x= –0 2 in tih.ee. ,givena xe2q –u a(t4ioan –, tbh)ex r+e q(4uair e–d 2 ebq +u act)i o=n 0 is. 0
oSolved Example # 3 The coefficient of x in the quadratic equation x2 + px + q = 0 was taken as 17 in place of 13, itse :
k roots were found to be – 2 and – 15. Find the roots of the original equation. n
e o
TSolution. Here q = (– 2) × (– 15) = 30, correct value of p = 13. Hence original equation is h
w.Self Praxc2t +ic 1e3 Pxr +o b3l0e m= s0 :a 1s. (x I+f 10,) ( xa r+e 3th) e= r0oots of the quadrroaotitcs eaqreu a –ti o1n0 ,a –x 23+ bx + c = 0 then find the quadratic P
w equation whose roots are al
p
: w (i) 2, 2 (ii)2,2 (iii) + 1,  + 1 (iv) 11, 11 (v)  ,  Bho
e ),
it (r1)2 b2 Sir
s2. If r be the ratio of the roots of the equation ax2 + bx + c = 0, show that = .
b r ac K.
e Ans.(1)(i) ax2 + 2bx + 4c = 0 (ii) a2x2 + (2ac – b2) x + c2 = 0 .
w (iii) ax2 – (2a – b) x + a + c – b = 0 (iv) (a + b + c)x2 – 2(a – c) x + a – b + c = 0 R
m (v) ac x2 – (b2 – 2ac) x + ac = 0 S.
3. Nature of Roots: (
o a
e fr Consider the quadratic equation, ax2 + bx + c = 0 having, as its roots; D b2 4 ac ariy
K
g
a R.
k g
c D = 0 D  0 a
a Roots are equal = = b/2a Roots are unequal h
P u
S
y :
d s
u h
t
t a
S M

d a, b, c  R & D > 0 a, b, c  R & D < 0 s,
a e
o Roots are real Roots are imaginary  = p + iq,  = p iq s
s
nl a
w Cl

o o
k
D e
a, b, c  Q & a, b, c  Q & T
E D is a perfect square D is not a perfect square
E  Roots are rational  Roots are irrational
R  i.e.  = p + q ,  = p  q
F
a = 1, b, c  & D is a perfect square
 Roots are integral.
Solved Example # 4: For what values of m the equation (1 + m)x2 – 2(1 + 3m)x + (1 + 8m) = 0 has equal roots.
Solution.
Given equation is (1 + m) x2 – 2(1 + 3m)x + (1 + 8m) = 0 ........(i)
Let D be the discriminant of equation (i).
Roots of equation (i) will be equal if D = 0.
or, 4(1 + 3m)2 – 4(1 + m) (1 + 8m) = 0
Successful People Replace the words like; "wish", "try" & "should" with "I Will". Ineffective People don't.
Get Solution of These Packages & Learn by Video Tutorials on www.MathsBySuhag.com
or, 4(1 + 9m2 + 6m – 1 – 9m – 8m2) = 0
or, m2 – 3m = 0 or, m(m – 3) = 0  m = 0, 3.
Solved Example # 5: Find all the integral values of a for which the quadratic equation (x – a) (x – 10) + 1 = 0 has
integral roots.
Solution.: Here the equation is x2 – (a + 10)x + 10a + 1 = 0. Since integral roots will always be rational it means D
m should be a perfect square.
o From (i) D = a2 – 20a + 96.
c  D = (a – 10)2 – 4  4 = (a – 10)2 – D 3
. If D is a perfect square it means we want difference of two perfect square as 4 which is possible only when (a –2
g
a 10)2 = 4 and D = 0. of
h  (a – 10) = ± 2  a = 12, 8 3
uSolved Example # 6: If the roots of the equation (x – a) (x – b) – k = 0 be c and d, then prove that the roots of thee
S equation (x – c) (x – d) + k = 0, are a and b. g
ySolution. By given condition pa
sB (Axb –o vae) s(xh o–w bs) t–h akt th (ex r–o oc)ts ( xo f– ( xd )– c) (xo –r (dx) –+ ck) = ( x0 –a rde) a+ aknd (bx. – a) (x – b)
hSelf Practice Problems : 1 .
at3. Let 4x2 – 4( – 2)x + – 2 = 0 ( R) be a quadratic equation. Find the value of  for which 88
M (i) Both roots are real and distinct. (ii) Both roots are equal. 8
5
w. (iii) Both roots are imaginary (iv) Both roots are opposite in sign. 0
(v) Both roots are equal in magnitude but opposite in sign. 3
w4. Find the values of a, if ax2 – 4x + 9 = 0 has integral roots. 9
w5. If P(x) = ax2 + bx + c, and Q(x) = – ax2 + dx + c, ac 0 then prove that P(x) . Q(x) = 0 has atleast two real roots.98
& Ans. (1) (i) (–, 2) (3,) (ii)  {2, 3} 0
(iii) (2, 3) (iv) (–, 2) (v) 
m 9,
1 1 7
o (2) a = , – 7
c 3 4 7
s.4. Common Roots: 3
0
e Consider two quadratic equations, a x2 + b x + c = 0 & a x2 + b x + c = 0. 9
s (i) If two quadratic equations 1have b1oth r1oots com2 mon,2 then 2the equation are identical and their3
s 0
a a b c 9
Cl co-efficient are in proportion. i.e. 1 = 1 = 1 . 0
a b c
o 2 2 2 e :
k c a  c a b c  b c n
e (ii) If only one root is common, then the common root '' will be:  = 1 2 2 1 = 1 2 2 1 o
T a b a b c a  c a h
1 2 2 1 1 2 2 1 P
w. Hence the condition for one common root is:
ww  c1a2  c2a12  c1a2  c2a1 opal
a1a1b2 a2b1 + b1a1b2 a2b1 + c1 = 0 Bh
:
ite  c1a2  c2a12 =a1b2 a2b1 b1c2  b2c1 Sir),
s
Note : If f(x) = 0 & g(x) = 0 are two polynomial equation having some common root(s) then those common root(s) is/are
b also the root(s) of h(x) = a f(x) + bg (x) = 0. K.
e
wSolved Example # 7: If x2 – ax + b = 0 and x2 – px + q = 0 have a root in common and the second equation has equalR.
m ap S.
roots, show that b + q = .
(
o 2 a
rSolution. Given equations are : x2 – ax + b= 0 and x2 – px + q = 0. y
e f Lroeotts boef ethqeu actoiomnm (1o)n a rroeot. ,Th eann dro tohtoss oef oefq euqautiaotnio (n2 )( 2w)i lal bree, an.d. Let be the other root of equation (1). ThusKari
ag Now  +  = a ........ (iii) R.
ck 2 == pb ................ ((ivv)) ag
a 2 = q ........ (vi) h
P L.H.S. = b + q =  + 2 = ( + ) ........ (vii) Su
y ap ()2 :
d and R.H.S. = = =  ( + ) ........ (viii) s
u 2 2 h
t
t from (7) and (8), L.H.S. = R.H.S. a
S M
Solved Example # 8: If a, b, c R and equations ax2 + bx + c = 0 and x2 + 2x + 9 = 0 have a common root, show that

d a : b : c = 1 : 2 : 9. s,
aSolution. Given equations are : x2 + 2x + 9 = 0 ........(i) e
o s
and ax2 + bx + c = 0 ........(ii) s
wnl Cmluesatr lbye r oimotasg oinf aeqryu aatniodn h (ei)n acree b imotahg rionoatrsy wsiinllc bee e cqoumatmioonn (.i) and (ii) have a common root, therefore common rootCla

o Therefore equations (i) and (ii) are identical o
k
D e
E  a = b = c  a : b : c = 1 : 2 : 9 T
1 2 9
E
Self Practice Problems : 6. If the equation x2 + bx + ac = 0 and x2 + cx + ab = 0 have a common root then
R
prove that the equation containing other roots will be given by x2 + ax + bc = 0.
F7. If the equations ax2 + bx + c = 0 and x3 + 3x2 + 3x + 2 = 0 have two common roots then show that
a = b = c.
a b c
8. If ax2 + 2bx + c = 0 and a x2 + 2b x + c = 0 have a common root and , , are in A.P. show that
1 1 1 a b c
1 1 1
a , b , c are in G.P.
1 1 1
Successful People Replace the words like; "wish", "try" & "should" with "I Will". Ineffective People don't.
Get Solution of These Packages & Learn by Video Tutorials on www.MathsBySuhag.com
5. Factorisation of Quadratic Expressions:
 The condition that a quadratic expression f(x) = ax2 + bx + c a perfect square of a linear expression, is Db2
 4 ac = 0.
 The condition that a quadratic expressionf(x, y)= ax²+2hxy + by² +2gx+ 2fy + c may be resolved into two linear
factors is that;
m
o a h g
c 3
g.  abc + 2fgh af² bg² ch² = 0 OR h b f = 0. 2
f
a g f c o
hSolved Example # 9: Determine a such that x2 – 11x + a and x2 – 14x + 2a may have a common factor. 4
uSolution. Let x – be a common factor of x2 – 11x + a and x2 – 14x + 2a. e
S Then x =  will satisfy the equations x2 – 11x + a = 0 and x2 – 14x + 2a = 0. g
y  2 – 11 + a = 0 and 2 – 14 + 2a = 0 pa
hsBSSoollu. Etixo.nS 1.o0Glv:iivnegn S( iqh) uoaawndd rt ha(iatii)tc b tehyxe cp erroexsspssr iemosnus iwlotiinpll l xibc2ea +tai o 2pn(ea mr f+ee cbtht +soq dcu,) awxr ee+ gi3fe (tbth ace =+d i2scca4r. i+m ainba) nwt iollf b ites ac oprererfsepcot nsdqiunagr eeq iuf aat i=o nb i=s zce.ro. 1 .
t 8
a i.e. 4(a + b + c)2 – 4.3 (bc + ca + ab) = 0 8
M or (a + b + c)2 – 3(bc + ca + ab) = 0 8
5
w. 1 0
or ((a – b)2 + (b – c)2 + (c – a)2) = 0 3
w 2 9
w which is possible only when a = b = c. 8
9
Self Practice Problems :
& 0
m 91.0. FIf oxr –wha bt ev aal fuaecst oorf cko tmhem eoxnp troe sasixo2n + ( b4 –x k+)cx 2a +n d2 (ak x+2 2+) xb +x 8+k c +pr 1o vwei ltl hbaet a (pae r–fe act) s=q uba r–e b ?. 9,
o11. If 3x2 + 2xy + 2y2 + 2ax – 4y + 11 can b1e resolved i2nto two2 linear factors, Prov1e tha2t is2 a ro1ot of the equation 7
c x2 + 4ax + 2a2 + 6 = 0. Ans. (1) 0, 3 77
s.6. Graph of Quadratic Expression: 3
0
e y = f (x) = ax2 + bx + c 9
s
3
s 0
a  D   b 2 9
Cl or y   = a x   0
o  4a  2a e :
k n
e  the graph between x, y is always a parabola. o
T h
 b D  P
w.  the coordinate of vertex are  , 
w  2a 4a al
w  If a > 0 then the shape of the parabola is concave upwards & if a < 0 then the shape of the parabola is op
 cthoen cpaavraeb doolwa ninwtaerrdsse.ct the yaxis at point (0, c). Bh
:
e  the xcoordinate of point of intersection of parabola with xaxis are the real roots of the quadratic ),
it equation f (x) = 0. Hence the parabola may or may not intersect the xaxis at real points. Sir
s
b7. Range of Quadratic Expression f (x) = a x2 + b x + c. K.
e (i) Absolute Range: .
w R
 D  .
m If a > 0  f (x)  , S
 4a  (
o a
r y
ge f a < 0  f (x) ,4Da Kari
a R.
D
ck Hence maximum and minimum values of the expression f (x) is4a in respective cases and it occurs ag
a h
P b u
S
y at x =2a (at vertex). :
d s
u (ii) Range in restricted domain: Given x [x1, x2] th
St b Ma
(a) If   [x , x ] then,
d 2a 1 2 s,
a      e
o f (x) min f(x1),f(x2) , max f(x1),f(x2) ss
nl a
w b Cl
(b) If   [x , x ] then,
o 2a 1 2 o
k
D e
E f (x) min f(x1),f(x2), D , max f(x1),f(x2), D  T
E   4a  4a
RSolved Example # 11 If c < 0 and ax2 + bx + c = 0 does not have any real roots then prove that
F (i) a – b + c < 0 (ii) 9a + 3b + c < 0.
Solution.
c < 0 and D < 0  f(x) = ax2 + bx + c < 0 for all x  R
 f(– 1) = a – b + c < 0
and f(3) = 9a + 3b + c < 0
Solved Example # 12 Find the maximum and minimum values of f(x) = x2 – 5x + 6.
Solution.
Successful People Replace the words like; "wish", "try" & "should" with "I Will". Ineffective People don't.
Get Solution of These Packages & Learn by Video Tutorials on www.MathsBySuhag.com
D b
minimum of f(x)= – at x = –
4a 2a
2524 5 1
= –   at x = = –
m  4  2 4
o  1 
c maximum of f(x) = Hence range is  ,. 3
.  4  2
g
f
a x2 x1 o
hSolved Example # 13 :Find the range of rational expression y = if x is real. 5
u x2 x1 e
ySSolution. y = x2 x1 pag
sB  (y – 1x)2x2 x+ (y1 + 1) x + y – 1 = 0
h  x is real  D 0 1 .
t 8
a 1  8
M  (y + 1)2 – 4(y – 1)2 0  (y – 3) (3y – 1)  0  y   , 3. 8
3  5
w. 0
wSolved Example # 14:Find the range of y = x2 , if x is real. 93
w 2x2 3x6 8
9
& x2 0
m Solution.: y = 2x2 3x6 9,
 2yx2 + 3yx + 6y = x + 2  2yx2 + (3y – 1) x + 6y – 2 = 0 7
o
 x is real 7
c 7
s.  D(3y –0 1)2 – 8y (6y – 2)  0  (3y – 1) (13y + 1) 0 03
e
9
s  1 1 3
s y   , . 0
a  13 3 9
ClSelf Practice Problems : 0

o12. If c > 0 and ax2 + 2bx + 3c = 0 does not have any real roots then prove that e :
k (i) a – 2b + 3c > 0 (ii) a + 4b + 12c > 0 n
Te (ab)2 ho
13. If f(x) = (x – a) (x – b), then show that f(x) – . P
w. 4
w14. For what least integral value of k the quadratic polynomial (k – 2) x2 + 8x + k + 4 > 0 x R. al
p
w x2 34x71 o
15. Find the range in which the value of function x2 2x7 lies x R. Bh
:
ebsite16. Fi nxd tRhe. interval in which 'm' lies so that the function y = m4xx2233xx4m can take all real values K. Sir),
w Ans. (14) k = 5. (15) (–, 5]  [9, ) (16) m  [1, 7] R.
8. Sign of Quadratic Expressions: .
m S
The value of expression, f (x) = ax2 + bx + c at x = x is equal to ycoordinate of a point on parabola (
e fro y>W =0e aagnxed2t +vsi ibxce xd i+fvf eecr rwseahn.ot pseo sxiticoonso orfd tinhaet ger iasp xh0 .w Hitehn rcees pife tchte t po0o xinta lxieiss aasb sohvoew thne. xaxis for some x = x0, then f (x0) Kariya
g
a R.
k g
c a
a h
P u
S
y :
d s
u h
t
t a
S M

d s,
a e
o s
s
nl a
w Cl

o o
k
D e
T
E
E
R
F NOTE:
(i)  x  R, y > 0 only if a > 0 & D  b² 4ac < 0 (figure 3).
(ii)  x  R, y < 0 only if a < 0 & D  b² 4ac < 0 (figure 6).
9. Solution of Quadratic Inequalities:
The values of 'x' satisfying the inequality, ax2 + bx + c > 0 (a 0) are:
(i) If D > 0, i.e. the equation ax2+ bx + c = 0 has two different roots <.
Then a > 0 x  () ()
Successful People Replace the words like; "wish", "try" & "should" with "I Will". Ineffective People don't.
Get Solution of These Packages & Learn by Video Tutorials on www.MathsBySuhag.com
a < 0  x  ()
(ii) If D = 0, i.e. roots are equal, i.e.  =.
Then a > 0 x  (, ) ()
a < 0  x 
(iii) If D < 0, i.e. the equation ax2+ bx + c = 0 has no real root.
m
Then a > 0  x  R
o a < 0  x  
c 3
. P(x) Q(x) R(x).........  2
g (iv) Inequalities of the form  0 can be quickly solved using the method of f
a A(x) B(x) C(x).........  o
h intervals, where A, B, C........, P, Q, R......... are linear functions of 'x'. 6
u e
S x2 6x7 g
ySolved Example # 15 Solve  2 a
hsBSolution. x2 – 6x +x 29 + 6 0x – 7x2 2x12 + 2 (x – 3)2 0  x  R 1 . p
at x2 x1 88
MSolved Example # 16: Solve > 0. 8
|x1| 5
w.Solution.  |x + 1| > 0 0
w  x  R – {–1} 93
w  x2 + x + 1 > 0  D = 1 – 4 = – 3 < 0 8
 x2 + x + 1 > 0  x  R  x  (–, – 1) (– 1,) 9
& 0
m x2 3x1 9,
oSolved Example # 17 x2 x1 < 3. 7
7
c 7
s.Solution. |x2 3x1| < 3. 03
e x2 x1 9
s  in x2 + x + 1 3
s D = 1 – 4 = – 3 < 0 0
a 9
Cl  x2 + x + 1 > 0  x  R  |x2 – 3x – 1| < 3(x2 + x + 1) 0
o  ((x42x 2– +3 x2 )– ( –1 )22 x–2 {–3 (6xx2 –+ 4x) + < 1 0)}2 < 0 e :
ek  (2x2 + 1) (x + 2) (x + 1) > 0  x  (–, – 2) (– 1,) on
TSelf Practice Problems : h
P
w.17. (i) |x2 + x | – 5 < 0 (ii) x2 – 7x + 12 < |x – 4|
w 2x 1 pal
w1189.. SSoollvvee thex i2neq9uatioxn2 (x2 + 3x + 1) (x2 +3x – 3) 5 Bho
:
bsite20. Find the value of parameter '' for which the inequality xx22xx11  3 is satisfied x R K. Sir),
e
w x2 5x4 R.
m 21. Solve x2 4  1 S.
(

e fro Ans. (17) (i) 1221,  2121 (ii) (2, 4) Kariya
ag (18) (–, – 3) (– 2, 3) (19) (–, – 4] [–2, –1]  [1,) R.
k  8 5  g
c (20) (–1, 5) (21) 0,    , a
a  5 2  h
P10. Location Of Roots: Su
y Let f(x) = ax²+ bx+ c, where a > 0 & a, b, c R. :
d s
u h
t
t a
S M

d s,
a e
o (i) (ii) (iii) s
s
nl a
w Cl

o (i) Conditions for both the roots of f (x) = 0 to be greater than a specified number‘x ’ are o
D b² 4ac 0; f(x ) > 0 & (b/2a) > x . 0 ek
(ii) Conditions for 0both the roots of0 f (x) = 0 to be smaller than a specified number ‘x ’ are T
E b² 4ac 0; f(x ) > 0 & (b/2a) < x . 0
E 0 0
(iii) Conditions for both roots of f(x) = 0 to lie on either side of the number ‘x ’ (in other words the number ‘x ’
R 0 0
lies between the roots of f(x) = 0), is f(x ) < 0.
F 0
(iv) (v)
Successful People Replace the words like; "wish", "try" & "should" with "I Will". Ineffective People don't.
Get Solution of These Packages & Learn by Video Tutorials on www.MathsBySuhag.com
(iv) Conditions that both roots of f (x) = 0 to be confined between the numbers x and
x , (x < x ) are b² 4ac 0; f(x ) > 0 ; f(x ) > 0 & x < (b/2a) < x . 1
2 1 2 1 2 1 2
(v) Conditions for exactly one root of f(x) = 0 to lie in the interval (x , x ) i.e.
1 2
x < x < x is f(x ).f(x ) < 0.
mEx.10.1 x2 – (m 1– 3) x + 2m = 01 2
o(a) Find values of m so that both the roots are greater than 2.
c 3
. 2
g
f
a o
h 7
u e
yS Condition - D 0  ((mm –– 13)) 2( m– 4 –m 9) 0 0 mm2 – (1–0m, +1 ]9 [ 90, ) ......(i) pag
hsB Condition - f(2) > 0  4 – (m – 3)2 + m > 0 m < 10...(ii), 1 .
Mat Condition - – 2ba > 2  m23 2  m > 7.....(iii) 888
5
w. Intersection of (i), (ii) and (iii) gives m [9, 10) Ans. 0
3
w(b) Find the values of m so that both roots lie in the interval (1, 2) 9
w 8
9

& 0

m 9,
7
o
7
s.c CCoonnddiittiioonn -- Df(1) >0 0  m1 – ( m(– – 3, )1 ]+ m [ 9>, 0)  4 > 0  m  R 03 7
e Condition - f(2) > 0  m < 10 9
s
3
s b m3 0
a Condition -V 1 < – < 2  1 < < 2  5 < m < 7 9
Cl 2a 2 0
o(c) iOnnteer sroeoctti oisn g greivaetes rm than 2 and oAthnesr. smaller than 1 e :
k n
e o
T h
P
w.
w al
p
w o
: CCoonnddiittiioonn -- ff((12)) << 00  4m < > 0 10  m  Bh
ite Intersection gives m   Ans. Sir),
s
b(d) Find the value of m for which both roots are positive. K.
e
.
w R
.
m S
(

o a
e fr CCoonnddiittiioonn -- Df(0) >0 0  mm > 0(–, 1] [9, ) Kariy
kag Condition - 2ba> 0  m23 > 0  m > 3 g R.
c intersection gives m [9,) Ans. a
a h
P(e) Find the values of m for which one root is (positive) and other is (negative). u
S
y :
d s
u h
t
t a
S M

d Condition - f(0) < 0  m < 0 Ans. s,
a e
(f) Roots are equal in magnitude and opposite in sign.
o s
sum of roots = 0  m = 3 s
wnl and fm(0) < 0 Ans.  m < 0 Cla
oEx.10.2 Find all the values of 'a' for which both the roots of the equation ko
D (a – 2)x2 + 2ax + (a + 3) = 0 lies in the interval (– 2, 1). e
E Sol. Case - T
E
R
F
When a – 2 > 0
 a > 2
Condition - f(–2) > 0 (a – 2)4 – 4a + a + 3 > 0  a – 5 > 0  a > 5
Successful People Replace the words like; "wish", "try" & "should" with "I Will". Ineffective People don't.
Get Solution of These Packages & Learn by Video Tutorials on www.MathsBySuhag.com
1
Condition - f(1)> 0  4a + 1 > 0  a > –
4
Condition - D 0  4a2 – 4(a + 3) (a – 2) 0  a 6
m b 2(a1)
Condition -V – <1 > 0  a (–, 1) (4, )
o 2a a2
c b 2a a4 3
g. Condition -V – 2< – 2a  2(a2) > – 2  a2 > 0 f 2
a o
h Intersection gives a (5, 6]. Ans. 8
uCase-when a – 2 < 0 e
S a < 2 g
y Condition - f(–2) < 0  a < 5 pa
hsB Condition - f(1) < 0,  a < – 41 1 .
t 8
a b 8
M Condition - – 2 < – < 1  a (–, 1) (4, ) 8
2a 5
w. Condition -V D 0  a 6 0
3
w  1 9
w intersection gives a  ,  8
 4 9

& 0
m complete solution is a , 1  (5, 6] Ans. 9,
 4 7
o
7
cSelf Practice Problems : 7
s.22. Let 4x2 – 4( – 2)x + – 2 = 0 ( R) be a quadratic equation find the value of  for which 3
(a) Both the roots are positive (b) Both the roots are negative 0
e
s (c) Both the roots are opposite in sign. (d) Both the roots are greater than 1/2. 9
3
s (e) Both the roots are smaller than 1/2. 0
a (f) One root is small than 1/2 and the other root is greater than 1/2. 9
Cl Ans. (a) [3,) (b)  (c) (–, 2) (d)  (e) (–, 2] (f) (3,) 0

o23. Find the values of the parameter a for which the roots of the quadratic equation e :
k x2 + 2(a – 1)x + a + 5 = 0 are n
e (i) positive (ii) negative (iii) opposite in sign. o
T Ans. (i) (–5, – 1] (ii) [4,) (iii) (–, – 5) Ph
w.24. Find the values of P for which both the roots of the equation
w 4x2 – 20px + (25p2 + 15p – 66) = 0 are less than 2. al
w Ans. (–, –1) op
25. Find the values of  for which 6 lies between the roots of the equation x2 + 2( – 3)x + 9 = 0. Bh
ite: Ans. ,34. Sir),
s
b26. Let 4x2 – 4( – 2)x + – 2 = 0 ( R) be a quadratic equation find the value of  for which K.
we(i) Exactly one root lies in 0, 1. (ii) Both roots lies in 0, 1. R.
m  2  2 S.
(
o  1 a
e fr(iii) AAnt lse.ast (oi)n (e– roo, 2t )lies (i3n,0),2.((iivi)) One root is g(iriei)a (t e–r th, a2n) 1/ 2(3 a,nd) other root is(i vsm)aller than 0. Kariy
ag27. Ixn2 w– h2aatx i n+t ear2v –a l1 m =u s0t ltihees nbuemtwbeeern ' a–' v2a aryn ds o4 .that both rooAtnss o.f the(– e 1q,u 3a)tion R.
ck28. Find the values of k, for which the quadratic expression ax2 + (a – 2) x – 2 is negative for exactly two integralag
a values of x. Ans. [1, 2) h
P11. Theory Of Equations: u
S
dy If1,2,3,......n are the roots of the equation; s :
u f(x) = a xn + a xn-1 + a xn-2 +.... + a x + a = 0 where a a .... a are all real & a  0 then,th
t 0 1 2 n-1 n 0, 1, n 0 a
S M
a a a a
ad 1 =a01 ,12 = +a02 ,123 =a03 ,.....,123........n = (1)na0n es,
o NOTE: s
s
nl (i) If  is a root of the equation f(x) = 0, then the polynomial f(x) is exactlydivisible by (x) or (x) is aa
w factor of f(x) and conversely. Cl
o (ii) Every equation of nth degree (n1) has exactly n roots & if the equation has more than n roots, it is ano
D identity. ek
(iii) If the coefficients of the equation f(x) = 0 are all real and + i is its root, then i is also a root. i.e.T
E imaginary roots occur in conjugate pairs.
E (iv) An equation of odd degree will have odd number of real roots and an equation of even degree will have
R even numbers of real roots.
F (v) If the coefficients in the equation are all rational &  +  is one of its roots, then
  is also a root where,  Q & is not a perfect square.
(vi) If there be any two real numbers 'a' & 'b' such that f(a) & f(b) are of opposite signs, then
f(x) = 0 must have odd number of real roots (also atleast one real root) between 'a' and 'b'.
(vii) Every equation f(x) = 0 of degree odd has atleast one real root of a sign opposite to that of its
last term. (If coefficient of highest degree term is positive).
Successful People Replace the words like; "wish", "try" & "should" with "I Will". Ineffective People don't.
Get Solution of These Packages & Learn by Video Tutorials on www.MathsBySuhag.com
Ex.11.1 2x3 + 3x2 + 5x + 6 = 0 has roots , ,  then find  +  + ,  + +  and .
3 5 6
 +  +  = = – +  +  = ,  = – = – 3.
2 2 2
Ex.11.2 Find the roots of 4x3 + 20x2 – 23x + 6 = 0. If two roots are equal.
m Let roots be,  and
o 20
c   +  +  = – 4 3
. 2
g  2 +  = – 5 .............(i)
f
a o
h   .  +  +  = – 23 9
u 4 e
S g
y  2 + 2 = – 23 & 2 = – 6 pa
sB from equation (i) 4 4
h 23 1 .
at 2 + 2 (– 5 – 2) = – 88
M 4 8
w.  2 – 10 – 42 = – 23  122 + 40 – 23 = 0 0 5
w 4 93
w 23 1 8
  = 1/2, – when  = 9
& 6 2 0
m from equation (i) 2 = 1 (– 5 – 1) = – 3 9,
o 4 2 7
7
c 23 7
s. when  = – 6 03
e
9
ss 2 = 2323 52x23  – 3 03
a 36   6  2 9
Cl 0
o   = 1,  = – 6 e :
k 2 n
Te Hence roots of equation = 1, 1, – 6 Ans. Pho
ww.Self Practice Problems : 2 2 al
29. Find the relation between p, q and r if the roots of the cubic equation x3 – px2 + qx – r = 0 are such that they are p
w in A.P. Ans. 2p3 – 9pq + 27r = 0 o
30. If ,, are the roots of the cubic x3 + qx + r = 0 then find the equation whose roots are Bh
e: (a)  + ,  + ,  +  Ans. x3 + qx – r = 0 ),
sit ((bc)) 2,,2,,2 AAnnss.. xx33 –+ q2xq2x –2 +r2 q =2 0x – r2 = 0 Sir
b (d) 3, 3, 3 Ans. x3 + 3x2r + (3r2 + q3) x + r3 = 0 K.
e
.
w R
SHORT REVISION
.
m S
(

o a
r y
e fThe general form of a quadratic equation in x is, ax2+ bx+ c = 0 , where a, b, c R & a 0. ari
gRESULTS : K
a R.
k b b24ac g
c1. The solution of the quadratic equation , ax² + bx + c = 0 is given by x = a
a 2a h
P The expression b2 – 4ac= D is called the discriminant of the quadratic equation. u
S
y 2. If  & are the roots of the quadratic equation ax² + bx + c = 0, then; :
ud (i) + = – b/a (ii) = c/a (iii)– = D/a. hs
t3. NATURE OF ROOTS: at
S (A) Consider the quadratic equation ax²+ bx+ c = 0 where a, b, c R & a 0 then ; M
d (i) D > 0  roots are real & distinct (unequal). s,
a (ii) D = 0  roots are real & coincident (equal). e
o (iii) D < 0  roots are imaginary . ss
nl (iv) If p+iq is one root of a quadratic equation, then the other must be the a
w Cl
conjugate piq & vice versa. (p, q R & i = 1).
o (B) Consider the quadratic equation ax2+ bx + c = 0 where a, b, c Q & a 0 then; ko
D (i) If D > 0 & is a perfect square , then roots are rational & unequal. e
T
E
(ii) If  = p + q is one root in this case, (where p is rational & q is a surd)
E
R then the other root must be the conjugate of it i.e. = p q & vice versa.
F
4. A quadratic equation whose roots are & is (x)(x) = 0 i.e.
x2(+)x+ = 0 i.e. x2(sum of roots)x + product of roots = 0.
5. Remember that a quadratic equation cannot have three different roots & if it has, it becomes an identity.
6. Consider the quadratic expression , y = ax²+ bx + c, a 0 & a, b, c R then ;
(i) The graph between x,y is always a parabola . If a > 0 then the shape of the
parabola is concave upwards & if a < 0 then the shape of the parabola is concave downwards.
Successful People Replace the words like; "wish", "try" & "should" with "I Will". Ineffective People don't.
Get Solution of These Packages & Learn by Video Tutorials on www.MathsBySuhag.com
(ii)  x R , y > 0 only if a > 0 & b² 4ac < 0 (figure 3) .
(iii)  x R , y < 0 only if a < 0 & b² 4ac < 0 (figure 6) .
Carefully go through the 6 different shapes of the parabola given below.
7. SOLUTION OF QUADRATIC INEQUALITIES:
ax2 + bx + c > 0 (a 0).
m (i) If D > 0, then the equation ax2+ bx + c = 0 has two different roots x < x .
1 2
o Then a > 0  x (, x1) (x2,)
c a < 0  x (x , x) 3
g. (ii) If D = 0, then roots are equal, i.e. 1x1 =2 x2. f 2
ha In that case aa >< 00 x x (, x1) (x1,) 0 o
1
u P(x)
S (iii) Inequalities of the form 0 can be quickly solved using the method of intervals. e
Q(x) g
y a
B8. MAXIMUM & MINIMUM VALUE of y = ax²+ bx + c occurs at x =(b/2a) according as ; p
aths a < 0 or a > 0 . y4ac4a b2 , if a > 0 & y, 4ac4a b2if a < 0 . 81 .
M9. COMMON ROOTS OF 2 QUADRATIC EQUATIONS [ONLY ONE COMMON ROOT] : 8
8
w. Let be the common root of ax²+ bx + c = 0 & ax2+ bx + c = 0 . Therefore 5
2  1 0
w a²+ b+ c = 0 ; a² + b+ c = 0. By Cramer’s Rule   93
w bcbc acac abab 8
& Therefore, = caca  bcbc. 0 9
m abab acac ,
So the condition for a common root is (ca ca)² = (ab ab)(bc bc). 9
o 7
c10. The condition that a quadratic function f(x,y) = ax²+ 2hxy + by² + 2gx + 2fy + c may be resolved into two linear factors is7
s. that ; 3 7
e a h g 0
ss abc + 2fgh af2 bg2 ch2 = 0 OR h b f = 0. 3 9
a 0
g f c 9
Cl11. THEORY OF EQUATIONS : 0
ko Iff(x)1 ,= 2a,0x3,n .+... ..a1nx na-r1e +th ea r2oxont-s2 o+f t.h..e. e q+u aatino-1nx; + an = 0 where a0, a1, .... an are all real & a0  0 then,ne :
e o
w.T 1 =aa01 ,12 = + aa02 , 123 = aa03 , .....,123........n = (1)n aa0n Ph
wNote : (i) Ifis a root of the equation f(x) = 0, then the polynomial f(x) is exactly divisible by (x) or (x) is a factor off(x)al
w and conversely . p
o
(ii) Every equation of nth degree (n1) has exactly n roots & if the equation has more than n roots, it is an identity. h
: (iii) If the coefficients of the equation f(x) = 0 are all real and + i is its root, then iis also a root. i.e.imaginary rootsB
te occur in conjugate pairs. r),
bsi (iv) If the coefficients in the equation are all rational & +  is one of its roots, then is also a root where ,. Si
e Q & is not a perfect square. K

w (v) If there be any two real numbers 'a' & 'b' such thatf(a) & f(b) are of opposite signs, then f(x) = 0 must have atleastR.
one real root between 'a' and 'b' .
m (vi) Every equation f(x) = 0 of degree odd has atleast one real root of a sign opposite to that of its last term. S.
o12. LOCATION OF ROOTS : Let f(x) = ax2+ bx+ c, where a > 0 & a, b, c R. a (
e fr (i) Cb2on 4daictio n0;s ff(od)r >b 0o t&h (thbe/2 ar)o >o tds. of f (x) = 0 to be greater than a specified number ‘d’ areariy
g (ii) Conditions for both roots of f(x) = 0 to lie on either side of the number ‘d’ (in other words the number ‘d’ liesK
a between the roots of f(x) = 0) is f(d) < 0. R.
k (iii) Conditions for exactly one root of f(x) = 0 to lie in the interval (d,e) i.e. d < x < e are b2 4ac > 0 & f(d) .f(e) <
c g
a 0. a
P (iv) Conditions that both roots of f (x) = 0 to be confined between the numbers p & q areuh
(p < q). b2 4ac 0; f(p) > 0; f(q) > 0 & p < (b/2a) < q. S
dy13. LOGARITHMIC INEQUALITIES :
(i) For a > 1 the inequality 0 < x < y & log x < log y are equivalent. s
u a a h
St ((iiii)i) FIfo ra >0 1< a t h<e 1n tlhoeg ixn e<q puality 0 < x <0 y < & x < lo agpax > logay are equivalent. Mat
ad ((viv)) IIff a0 >< 1a < t h1e tnh e lno g laaoxg >x p < p  xx > > a app es,
o (vi) If 0 < a < 1 then logax > p  0 < x < ap s
l a s
n EXERCISE–1 a
wQ.1 If the roots of the equation [1/(x+p)] + [1/(x+q)] = 1/r are equal in magnitude but opposite in sign, show that p+q = 2r & thatCl
o the product of the roots is equal to (1/2)(p2+q2). o
D k
e
E Q.2 If x2 x cos(A + B) + 1 is a factor of the expression, T
2x4 + 4x3 sin A sin B x2(cos 2A + cos 2B) + 4x cos A cos B 2. Then find the other factor.
E
Q.3 , are the roots of the equation K(x2 –x) + x+ 5 = 0. If K & K are the two values of K for which the roots, are connected
R 1 2
by the relation (/) + (/) = 4/5. Find the value of
F (K/K )+(K/K ).
1 2 2 1
Q.4 If the quadratic equations, x2 + bx + c = 0 and bx2 + cx + 1 = 0 have a common root then prove that either b + c + 1 = 0 or b2 + c2
+ 1 = bc + b + c.
 p2  p2
Q.5 If the roots of the equation 1q x2p(1q)xq(q1) = 0 are equal then show that
 
2 2
 
Successful People Replace the words like; "wish", "try" & "should" with "I Will". Ineffective People don't.